Question

$$ax+by=c$$

Solve for a (complex solution)

$\left\{\begin{matrix}a=-\frac{by-c}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&c=by\text{ and }x=0\end{matrix}\right.$

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Solve for b (complex solution)

$\left\{\begin{matrix}b=-\frac{ax-c}{y}\text{, }&y\neq 0\\b\in \mathrm{C}\text{, }&c=ax\text{ and }y=0\end{matrix}\right.$

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Solve for a

$\left\{\begin{matrix}a=-\frac{by-c}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&c=by\text{ and }x=0\end{matrix}\right.$

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Solve for b

$\left\{\begin{matrix}b=-\frac{ax-c}{y}\text{, }&y\neq 0\\b\in \mathrm{R}\text{, }&c=ax\text{ and }y=0\end{matrix}\right.$

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